The series $\sum_{n=1}^{\infty} \frac {\cos n}{2n^{\alpha}}$ converges for $\alpha \in (0,1)$.

I'm trying to solve this problem: Let $$\alpha>0$$ and $$a_{n}=\frac{\cos n}{2n^{\alpha}}$$ for all $$n\in\mathbb{N}$$. Prove that the series $$\sum_{n=1}^{\infty}a_{n}$$ coneverges.

For $$\alpha>1$$, we have that $$\mid\cos x\mid\leq1$$ for all $$x\in\mathbb{R}$$ and therefore for all $$n\in\mathbb{N}$$: $$\left\lvert \frac{\cos n}{2n^{\alpha}}\right\rvert=\frac{\lvert\cos n\rvert}{2n^{\alpha}}\leq\frac{1}{2}\cdot\frac{1}{n^{\alpha}}$$

Using the fact that $$\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}$$ is convergent for $$\alpha>1$$, we have for the comparison test that $$\sum_{n=1}^{\infty}a_{n}$$ is absolutely convergent for $$\alpha>1$$. And with this $$\sum_{n=1}^{\infty}a_{n}$$ is convergent. But I don't know how can I prove that $$\sum_{n=1}^{\infty}a_{n}$$ converges when $$0<\alpha<1$$. Could you help me or give me some idea to prove this?

• Are you sure it converges? – herb steinberg Aug 19 '19 at 19:58
• Use an Abel summation, knowing that the partial sums of $\cos{n}$ are bounded. – Mindlack Aug 19 '19 at 19:59

In the following, we derive Abel's test and Dirichlet's test for testing the convergence of infinite series in the form $$\sum_{n=1}^{\infty}a_{n}b_{n}$$. Firstly, we introduce Abel's transform, which is the discrete version of integration-by-part. Then we apply it to the difference of partial sums $$S_{n+k}-S_{n}$$.

Section 1: Abel's transform. Let $$(a_{n})$$ and $$(b_{n})$$ be sequences of real numbers. For each $$n\in\mathbb{N}$$, let $$B_{n}=\sum_{k=1}^{n}b_{k}$$. Define $$B_{0}=0$$. Let $$N\in\mathbb{N}$$. We rewrite the partial sum $$\sum_{n=1}^{N}a_{n}b_{n}$$ as $$\begin{eqnarray*} \sum_{n=1}^{N}a_{n}b_{n} & = & \sum_{n=1}^{N}a_{n}(B_{n}-B_{n-1})\\ & = & \sum_{n=1}^{N}a_{n}B_{n}-\sum_{n=1}^{N}a_{n}B_{n-1}\\ & = & \sum_{n=1}^{N}a_{n}B_{n}-\sum_{n=1}^{N-1}a_{n+1}B_{n}\\ & = & a_{N}B_{N}-\sum_{n=1}^{N-1}(a_{n+1}-a_{n})B_{n}. \end{eqnarray*}$$ This is known as Abel's transform. Comparing to the well-known integration-by-part formula $$\int_{a}^{b}f(x)g'(x)dx=[f(x)g(x)]_{a}^{b}-\int_{a}^{b}g(x)f'(x)dx,$$ we note that $$(a_{n})$$ is analog to $$f$$ while $$(b_{n})$$ is analog to $$g'$$.

Section 2: Define partial sum $$S_{n}=\sum_{k=1}^{n}a_{k}b_{k}$$. Let $$n,k\in\mathbb{N}$$. We consider $$S_{n+k}-S_{n}$$. Applying Abel's transform, we have $$\begin{eqnarray*} S_{n+k}-S_{n} & = & \sum_{i=n+1}^{n+k}a_{i}b_{i}\\ & = & a_{n+k}(B_{n+k}-B_{n})-\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})(B_{i}-B_{n}). \end{eqnarray*}$$ Now, we are ready to state and prove the following two theorems.

Theorem 1: If $$(a_{n})$$ is monotone and bounded, and the series $$\sum_{n=1}^{\infty}b_{n}$$ is convergent, then the series $$\sum_{n=1}^{\infty}a_{n}b_{n}$$ is convergent.

Proof: Choose $$M>0$$ such that $$|a_{n}|\leq M$$ for all $$n$$. Let $$\varepsilon>0$$ be arbitrary. Choose $$N$$ such that $$|B_{n+k}-B_{n}|<\varepsilon$$ whenever $$n\geq N$$ and $$k\in\mathbb{N}$$. For any $$n\geq N$$ and $$k\in\mathbb{N}$$, we have that: $$\begin{eqnarray*} |S_{n+k}-S_{n}| & \leq & |a_{n+k}(B_{n+k}-B_{n})|+\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}||B_{i}-B_{n}|\\ & \leq & M\varepsilon+\sum_{i=n+1}^{n+k-1}\varepsilon|a_{i+1}-a_{i}|\\ & \leq & M\varepsilon+\varepsilon|\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})|\\ & \leq & M\varepsilon+\varepsilon|a_{n+k}-a_{n+1}|\\ & \leq & 3M\varepsilon \end{eqnarray*}$$ This shows that $$(S_{n})$$ is a Cauchy sequence and hence $$\sum_{n=1}^{\infty}a_{n}b_{n}$$ is convergent. In the above, we have used the fact that $$(a_{n})$$ is monotone to obtain $$\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}|=|\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})|$$.

Theorem 2: If $$(a_{n})$$ is monotone and $$a_{n}\rightarrow0$$, and $$(B_{n})$$ is bounded (where $$B_{n}:=\sum_{k=1}^{n}b_{k}$$), then the series $$\sum_{n=1}^{\infty}a_{n}b_{n}$$ is convergent.

Proof: Let $$M>0$$ be such that $$|B_{n}|\leq M$$ for all $$n$$. Let $$\varepsilon>0$$ be arbitrary. Choose $$N$$ such that $$|a_{n}|<\varepsilon$$ whenever $$n\geq N$$. For any $$n\geq N$$ and $$k\in\mathbb{N}$$, we have: $$\begin{eqnarray*} |S_{n+k}-S_{n}| & \leq & |a_{n+k}(B_{n+k}-B_{n})|+\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}||B_{i}-B_{n}|\\ & \leq & 2M\varepsilon+2M\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}|\\ & = & 2M\varepsilon+2M|a_{n+k}-a_{n+1}|\\ & \leq & 6M\varepsilon. \end{eqnarray*}$$ Therefore $$(S_{n})$$ is a Cauchy sequence and hence the series $$\sum_{n=1}^{\infty}a_{n}b_{n}$$ is convergent.

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Now, we go to prove that for each $$\alpha\in(0,1)$$, the series $$\sum_{n=1}^{\infty}\frac{\cos n}{2n^{\alpha}}$$ is convergent.

Proof: Let $$a_{n}=\frac{1}{2n^{\alpha}}$$, $$b_{n}=\cos n$$. Clearly $$(a_{n})$$ is monotonic decreasing and $$a_{n}\rightarrow0$$. Let $$B_{n}=\sum_{k=1}^{n}b_{k}$$. We go to show that $$(B_{n})$$ is bounded, then apply Theorem 2...

We go to compute $$\sum_{k=0}^{n-1}\cos k\theta$$ and demonstrate the use of complex numbers. Let $$\theta\in\mathbb{R}$$ be such that $$\cos\theta\neq1$$. Define $$z=\cos\theta+i\sin\theta$$. By DeMoivre Theorem, $$z^{k}=\cos(k\theta)+i\sin(k\theta)$$ for any $$k\in\mathbb{N}$$. On one hand, $$\begin{eqnarray*} 1+z+z^{2}+\ldots+z^{n-1} & = & \frac{1-z^{n}}{1-z}\\ & = & \frac{1-(\cos n\theta+i\sin n\theta)}{1-(\cos\theta+i\sin\theta)}\\ & = & \frac{2\sin^{2}(\frac{n\theta}{2})-2i\sin(\frac{n\theta}{2})\cos(\frac{n\theta}{2})}{2\sin^{2}(\frac{\theta}{2})-2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}\\ & = & \frac{\sin(\frac{n\theta}{2})\left[\sin(\frac{n\theta}{2})-i\cos(\frac{n\theta}{2})\right]}{\sin(\frac{\theta}{2})\left[\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})\right]}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\frac{\left[\sin(\frac{n\theta}{2})-i\cos(\frac{n\theta}{2})\right](i)}{\left[\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})\right](i)}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\frac{\cos(\frac{n\theta}{2})+i\sin(\frac{n\theta}{2})}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\left[\cos(\frac{n\theta}{2}-\frac{\theta}{2})+i\sin(\frac{n\theta}{2}-\frac{\theta}{2})\right]. \end{eqnarray*}$$ On the other hand, $$\begin{eqnarray*} 1+z+z^{2}+\ldots+z^{n-1} & = & \sum_{k=0}^{n-1}\cos k\theta+i\sum_{k=1}^{n-1}\sin k\theta. \end{eqnarray*}$$ Comparing the real part, we obtain: $$\sum_{k=0}^{n-1}\cos k\theta=\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cos\left(\frac{(n-1)\theta}{2}\right).$$ If $$\cos\theta=1$$, then $$\theta=2m\pi$$ for some $$m\in\mathbb{Z}$$. It follows that $$\cos k\theta=1$$ for all $$k\in\mathbb{N}$$. Therefore $$\sum_{k=0}^{n-1}\cos k\theta=n.$$ We conclude that: If $$\theta\in\mathbb{R}$$ such that $$\cos\theta\neq1$$, then $$|\sum_{k=0}^{n-1}\cos k\theta|\leq\left|\frac{1}{\sin(\frac{\theta}{2})}\right|.$$ Hence, the partial sum of the infinite series $$1+\cos\theta+\cos2\theta+\ldots$$ is bounded by $$\left|\frac{1}{\sin(\frac{\theta}{2})}\right|$$. Clearly $$\cos1\neq1$$, so $$(B_{n})$$ is bounded.

Hint

Dirichlet’s test may be your friend.